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3 years ago

13

What is the maximum volume of 0.25 M sodium hypochlorite solution (NaOCl, laundry bleach) that can be prepared by dilution of 1.

00 L of 0.80 M NaOCl?

2 answers:

ehidna [41]3 years ago

8 0

Answer:

0.313L

Explanation:

0.25Mx1.00L=0.80Mx V2

so, 0.25x1.00/0.80=0.313L

Iteru [2.4K]3 years ago

7 0

Answer: 3.2L

Explanation:

C1 = 0.80M

V1 = 1L

C2 = 0.25M

V2 =?

C1V1 = C2V2

0.8 x 1 = 0.25 x V2

V2 = (0.8 x 1) /0.25 = 3.2L

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A gas occupying a volume of 656.0 mL at a pressure of 0.884 atm is allowed to expand at constant temperature until its pressure

7nadin3 [17]

Answer:

1.14 × 10³ mL

Explanation:

Step 1: Given data

Initial volume of the gas (V₁): 656.0 mL

Initial pressure of the gas (P₁): 0.884 atm

Final volume of the gas (V₂): ?

Final pressure of the gas (P₂): 0.510 atm

Step 2: Calculate the final volume of the gas

If we assume ideal behavior, we can calculate the final volume of the gas using Boyle's law.

P₁ × V₁ = P₂ × V₂

V₂ = P₁ × V₁/P₂

V₂ = 0.884 atm × 656.0 mL/0.510 atm = 1.14 × 10³ mL

5 0

3 years ago

Suppose a 2.95 g of potassium iodide is dissolved in 350. mL of a 62.0 m M aqueous solution of silver nitrate. Calculate the fin

STALIN [3.7K]

Answer : The final molarity of iodide anion in the solution is 0.0508 M.

Explanation :

First we have to calculate the moles of $KI$ and $AgNO_3$ .

$\text{Moles of }KI=\frac{\text{Mass of }KI}{\text{Molar mass of }KI}$

Molar mass of KI = 166 g/mole

$\text{Moles of }KI=\frac{2.95g}{166g/mole}=0.0178mole$

and,

$\text{Moles of }AgNO_3=\text{Concentration of }AgNO_3\times \text{Volume of solution}=0.0620M\times 0.350L=0.0217mole$

Now we have to calculate the limiting and excess reagent.

The given chemical reaction is:

$KI+AgNO_3\rightarrow KNO_3+AgI$

From the balanced reaction we conclude that

As, 1 mole of KI react with 1 mole of $AgNO_3$

So, 0.0178 mole of KI react with 0.0178 mole of $AgNO_3$

From this we conclude that, $AgNO_3$ is an excess reagent because the given moles are greater than the required moles and KI is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $AgI$

From the reaction, we conclude that

As, 1 mole of $KI$ react to give 1 mole of $AgI$

So, 0.0178 moles of $KI$ react to give 0.0178 moles of $AgI$

Thus,

Moles of AgI = Moles of $I^-$ anion = Moles of $Ag^+$ cation = 0.0178 moles

Now we have to calculate the molarity of iodide anion in the solution.

$\text{Concentration of }AgNO_3=\frac{\text{Moles of }AgNO_3}{\text{Volume of solution}}$

$\text{Concentration of }AgNO_3=\frac{0.0178mol}{0.350L}=0.0508M$

Therefore, the final molarity of iodide anion in the solution is 0.0508 M.

3 0

3 years ago

Which of the following statements is true?

eduard

Answer:

Lines of latitude are used to divide Earth into 24 time zones

Explanation:

5 0

3 years ago

Density is the _____ of a certain ______

olga2289 [7]

I think the first blank is "solidity" and second is "object."

3 0

4 years ago

What is the mass of 2.16 moles of sulfur dioxide (SO2)?

Luda [366]

<h3>Answer:</h3>

138 g SO₂

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 2.16 moles SO₂

[Solve] grams (mass) SO₂

<u>Step 2: Identify Conversions</u>

[PT] Molar Mass of S - 32.07 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of SO₂ - 32.07 + 2(16.00) = 64.07 g/mol

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 2.16 \ mol \ SO_2(\frac{64.07 \ g \ SO_2}{1 \ mol \ SO_2})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 138.391 \ g \ SO_2$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

138.391 g SO₂ ≈ 138 g SO₂

6 0

3 years ago